phc-discussions - Re: Dumb idea of the day: Public key crypto based on random permutations

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Message-ID: <CAOLP8p6svseUUMwGTmNCQcVrot2gFoDJ6Gtp+_UBz--EiUZX7w@mail.gmail.com>
Date: Sat, 15 Aug 2015 10:07:03 -0700
From: Bill Cox <waywardgeek@...il.com>
To: "discussions@...sword-hashing.net" <discussions@...sword-hashing.net>
Subject: Re: Dumb idea of the day: Public key crypto based on random permutations

Cripes!  I meant to reply to the list.  This system as I wrote it is
trivial to break.  Alice publishes m*A, where A is her secret, but Eve can
reverse the computation just doing Finv(A) - 1.

Samuel Neves has given me some extensive pointers and I have week or so of
reading to do :)

It is still interesting, at least to me, because other choices of F do seem
to create secure crypto systems.  In particular for Edwards curves, for d
== -1, the function F is = Sqrt[2] EllipticF[ArcSin[x], -1].  However, I
can't do the modular arithmetic on this!  An easier example is crypto on
the unit circle (setting d == 0).  In this case, the function F(x) =
ArcSin(x).  I still can't compute this with modular arithmetic.

Bill

On Sat, Aug 15, 2015 at 6:12 AM, Bill Cox <waywardgeek@...il.com> wrote:

> This is what my code from last night does... I think!  This is too simple
> and obvious to be new, yet too useful to work.  Can one of you guys debunk
> this quickly for me before I get too excited?  I coded it in 56 lines of
> attached Python, and it seems to work.  Attack away!
>
> In short, I think I figured out how to create simple and fast public key
> protocols based on the security of random permutations and no other
> assumptions.  If true, I think this would be a big deal.  I seems like it
> will be faster than elliptic curves, requiring no more bits, and also
> appears to be post-quantum secure when the random permutation is.  The
> algorithm is super-simple, easily in the realm of what we can prove secure.
>
> The following construction allows us to turn just about any random
> permutation from n bits to n bits into an addition operator, suitable for
> abelian group addition.  Let F(x) be a random permutation of n-bits, such
> as AES.  Define the @ operator as follows:
>
>     a @ b = Finv(F(a) + F(b))
>
> This seems to be real addition.  From Wikipedia, here are the 5 properties
> I have to prove to show that this creates an abelian group:
>
> - Closure
> Since F is a random permutation, there is a value o = Finv(1).  Consider
> the sequence:
>
>     a, a @ o, A @ o @ o, A @ o @ o @ o, ....
>
> This sequence goes through all combinations of F(a) + k, for any k, before
> applying Finv.  Since it's a random permutation, Finv also goes through all
> values.
>
> - Associativity
> Obvious from definition of a @ b
>
> - Identity element
> The identity element is i = Finv(0).  a @ i = Finv(F(a) + F(i)) =
> Finv(F(a) + F(Finv(0))) = Finv(F(a) + 0) = a
>
>  - Inverse element
> The inverse of element 'a' is Finv(-F(a)):
>
>     a @ -a = Finv(F(a) + F(-a)) = Finv(F(a) + F(Finv(-F(a)))) = Finv(F(a)
> - F(a)) = Finv(0) = o
>
> - Commutativity
> We have to show (a @ b) @ c = a @ (b @ c):
>
>     (a @ b) @ c = Finv(F(Finv(F(a) + F(b))) + F(c))
>         = Finv(F(a) + F(b) + F(c))
>         = Finv(F(a) + Finv(F(F(b) + F(c))))
>         = a @ (b @ c)
>
> That should prove it works.  However, just because it's an abelian group
> based on random permutations doesn't prove it's secure.  Where are the
> holes?
>
> Bill
>

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